# How do you use partial fractions to find the integral int x/(16x^4-1)dx?

Jul 24, 2017

The answer is $= \frac{1}{16} \ln \left(| 2 x + 1 |\right) + \frac{1}{16} \ln \left(| 2 x - 1 |\right) - \frac{1}{16} \ln \left(| 4 {x}^{2} + 1 |\right) + C$

#### Explanation:

We need

$\int \frac{\mathrm{dx}}{x} = \ln \left(| x |\right) + C$

Let's factorise the denominator

$16 {x}^{4} - 1 = \left(4 {x}^{2} - 1\right) \left(4 {x}^{2} + 1\right) = \left(2 x + 1\right) \left(2 x - 1\right) \left(4 {x}^{2} + 1\right)$

We can perform the decomposition into partial fractions

$\frac{x}{16 {x}^{4} - 1} = \frac{x}{\left(2 x + 1\right) \left(2 x - 1\right) \left(4 {x}^{2} + 1\right)}$

$= \frac{A}{2 x + 1} + \frac{B}{2 x - 1} + \frac{C x + D}{4 {x}^{2} + 1}$

$= \frac{A \left(2 x - 1\right) \left(4 {x}^{2} + 1\right) + B \left(2 x + 1\right) \left(4 {x}^{2} + 1\right) + \left(C x + D\right) \left(2 x + 1\right) \left(2 x - 1\right)}{\left(2 x + 1\right) \left(2 x - 1\right) \left(4 {x}^{2} +\right)}$

The denominators are the same, we compare the numerators

$x = A \left(2 x - 1\right) \left(4 {x}^{2} + 1\right) + B \left(2 x + 1\right) \left(4 {x}^{2} + 1\right) + \left(C x + D\right) \left(2 x + 1\right) \left(2 x - 1\right)$

Let $x = - \frac{1}{2}$, $\implies$, $- \frac{1}{2} = - 4 A$, $\implies$, $A = \frac{1}{8}$

Let $x = \frac{1}{2}$, $\implies$, $\frac{1}{2} = 4 B$, $\implies$, $B = \frac{1}{8}$

Let $x = 0$, $\implies$, $0 = - A + B - D$, $\implies$, $D = 0$

Coefficients of ${x}^{3}$,

$0 = 8 A + 8 B + 4 C$, $\implies$, $4 C = - 2$, $C = - \frac{1}{2}$

Therefore,

$\frac{x}{16 {x}^{4} - 1} = \frac{\frac{1}{8}}{2 x + 1} + \frac{\frac{1}{8}}{2 x - 1} + \frac{- \frac{1}{2} x}{4 {x}^{2} + 1}$

$\int \frac{x \mathrm{dx}}{16 {x}^{4} - 1} = \int \frac{\frac{1}{8} \mathrm{dx}}{2 x + 1} + \int \frac{\frac{1}{8} \mathrm{dx}}{2 x - 1} + \int \frac{- \frac{1}{2} x \mathrm{dx}}{4 {x}^{2} + 1}$

$= \frac{1}{16} \ln \left(| 2 x + 1 |\right) + \frac{1}{16} \ln \left(| 2 x - 1 |\right) - \frac{1}{16} \ln \left(| 4 {x}^{2} + 1 |\right) + C$